TITLE: Linear Algebra: determine whether the sets span the same subspace
QUESTION [15 upvotes]: So I am stuck on 51 here:

51. Determine whether the sets $S_1$ and $S_2$ span the same subspace of $\mathbb{R}^3$:
  $$\begin{align*}
S_1 &= \Bigl\{ (1,2,-1),\ (0,1,1),\ (2,5,-1)\Bigr\}\\
S_2 &= \Bigl\{ (-2,-6,0),\ (1,1,-2)\Bigr\}
\end{align*}$$

What I did to solve it was to multiply each vector in set $S_1$ by $C$, add the vectors together and set them to zero. Once I reduced the resulting matrix I got the following result:
$$\begin{align*}
c_1&=-2t\\
c_2&=-t\\
c_3&=t
\end{align*}$$
So this result would be linearly dependent. Then I did the same thing for the second set and I got that $c_1=c_2=0$ which means that it is linearly independent. In conclusion, I said that set $S_1$ and $S_2$ does not span the same subspace of $\mathbb{R}^3$. The book says they do. Could anyone point out where I went wrong?
Thanks.

REPLY [7 votes]: For this kind of problem, finding the reduced row echelon form (rref) seems to be a rather straightforward method.
For the set $S_1$, you form the matrix
$$
\begin{pmatrix}
  1 & 2 &-1 \\
  0 & 1 & 1 \\
  2 & 5 &-1
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 2 &-1 \\
  0 & 1 & 1 \\
  0 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 2 &-1 \\
  0 & 1 & 1 \\
  0 & 0 & 0
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 0 &-3 \\
  0 & 1 & 1 \\
  0 & 0 & 0
\end{pmatrix}
$$
For the set $S_2$, you form the matrix
$$
\begin{pmatrix}
 -2 &-6 & 0 \\
  1 & 1 &-2
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 3 & 0 \\
  1 & 1 &-2
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 3 & 0 \\
  0 &-2 &-2
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 3 & 0 \\
  0 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 0 &-3 \\
  0 & 1 & 1
\end{pmatrix}
$$
Since we know that elementary row operations do not change row space, we see that
$$\newcommand{\span}{\operatorname{span}}\span(S_1) = \span\{(1,0,-3),(0,1,1)\} = \span(S_2).$$
This methods works in general. If the non-zero rows in the rrefs are different, that means $\span(S_1) \ne \span(S_2)$.
To formulate this more clearly, we know that:

Theorem. Let $A$ and $B$ be two $m\times n$ matrices over the same field. The following conditions are equivalent:

The matrices $A$ and $B$ are row equivalent.
The matrices $A$ and $B$ have the same row space.
The matrices $A$ and $B$ have the same reduced row echelon form.